Negative (Downward) Launch - Online Tutor, Practice Problems & Exam Prep

Projectile motion can involve launching objects downwards, such as throwing a rock from a building. The initial vertical velocity (v_0y) is negative due to the chosen positive direction being upwards. To find the height of the building, use the equation for vertical displacement: Δy = v_0yt + 1/2a_yt², where a_y is the acceleration due to gravity (g = -9.8 m/s²). The absolute value of the displacement gives the building's height.

Solving Downward Launch Problems

Everyone, welcome back. So by now, we've seen how to solve projectile motion problems. And in some cases, instead of launching an object upwards or even horizontally, you may actually launch objects downwards. For example, if you threw a rock downwards from a building. Now don't worry because you're going to solve these problems the exact same way. It's just another variation of projectile motion. We'll use the same exact steps and equations to solve. The only thing that's different about these problems is that when you launch objects downwards, when you separate their initial velocity into the x and y axes, that \( v_0y \) will actually always be negative. It will always be. And it's basically just because of the way that we choose the upward direction to be positive. So let me just go ahead and walk you through this example. We'll see that it's really not so different from how we solve any other projectile motion problem. Let's get started.

So in this problem here, we have a rock that's thrown at 5 meters per second at an angle of 37 degrees downward. In other words, the \( v_0 \) here is 5, and the angle from the horizontal is 37 degrees downward. So because this angle here is actually pointing down and this angle is actually going to be clockwise from horizontal, this angle is actually going to be negative 37 degrees. That's important there. We're told that it hits 10 meters from the building, and we're going to calculate the height of the building. Let's go ahead and just stick to the steps here. First thing we're going to do is draw the path in the x and y and label the points of interest. In other words, this ball really in the horizontal is going to go all the way over here. The path is going to look something like this where it's going to sort of curve downward, and that means that the path in the y-axis really just goes from the building all the way down to the ground. So these points of interest in the past is actually pretty straightforward. So where the point of interest is really just the initial, we label that as A. Now there's no other sort of max height. It doesn't curve upwards or downwards or something like this. It only really just goes from the initial, and then it just hits the ground. So there's really only two points of interest in this problem, which makes setting up our interval and our pass pretty straightforward.

Okay, so now let's figure out the target variable. We're told a couple of other pieces of information here. One of them is that this thing lands at 10 meters from the building, which is really just a \( \Delta x \). Right? So \( \Delta x \) is 10. But that's not what we want. We want the height of the building, and that's really just this vertical displacement over here. So this is basically the full height the rock falls as you're going from the roof up down to the ground. So this is going to be my height, which is really just \( \Delta y \) in this problem. And so that's really what we're looking for. It's actually really going to be the absolute value of \( \Delta y \) because the height is always just a positive number.

Okay, so that's the target variable. What about our interval and our UAM equation? Well, if we're looking for \( \Delta y \) in this problem then we're actually just going to take a look at our interval from A all the way to B. That's from the rooftop all the way down to the ground. So we're really just trying to figure out \( \Delta y \) from A to B because we're actually just looking at the interval from A to B. So this is going to be A to B. That's my interval over here. So what are my variables? Well, I've got \( \Delta y \) from A to B. That's what I really what I'm trying to solve for. Well, let's list out the other variables. I need the initial velocity in the y direction. I don't have that but I can find it because I have the initial velocity and the angle. What about the final velocity? I don't know what the velocity is in the y direction when it hits the ground. What else? So I've got so let's see. I've got my \( a \) in the y direction, which is equal just to negative \( g \), which is just negative 9.8. And then what about the time, the time that it takes in the air? That's \( \Delta t \) from A to B. I actually don't know anything about the time either. So I've got some work to do here because I have to figure out 3 out of 5 variables. Alright? Let's start off with the initial velocity in the y-axis because we can actually figure that out pretty straightforward. So if I want to split this out and get the x and y velocities over here, that's really just going to be \( v_0 \) . It's actually just going to be \( v_{a\_y} \) sorry. \( v_{ax} \), which is really just going to be \( v_a \times \cos \) of this angle. So the cosine of the angle here is going to be 37 degrees. In other words, that's just going to be 5 times the cosine of negative 37. If you go ahead and plug this in, what you're actually going to get is you're actually going to get 4. And then if you do the same thing for the y-axis, your \( v_0y \), this is really just going to be \( v_a \cos(\theta) \). In other words, this is really just going to be 5 I'm sorry. This is sine. My bad. This is going to be the sine of that angle. This is really just going to be 5 times the sine of negative 37 degrees. And when you plug this into your calculator, what you're actually going to get is an initial velocity that is negative 3. That's exactly what I meant over here by your initial velocity always being negative when you actually split it up into its components. And what's really cool about this is that you never have to, like, remember to stick a negative sign. When you plug in magnitude and direction, this will always actually just give you the appropriate sign. So it's negative 3.

Alright, so this is going to be negative 3. You still don't know anything about the initial about the final velocity in the y-axis. So I really just need one other variable. It's either going to be the final velocity or the time. But one of the things I want you to remember here is that whenever you sort of get stuck in the x and y axis, you should always just go to the other axis to solve. So if you're using if you get stuck in an x-axis, solve it with a y-axis equation and then vice versa. So we got stuck here with the y-axis. This is our y-axis equations. So what I'm going to do here is to figure out the time, I'm going to go ahead and look at the x-axis. So if I go to the x-axis over here, there's really only 3 variables because I just have \( \Delta x = v_x t \). Okay? So in other words, I've got \( \Delta x = v_x t \). And so, really, what happens is I know what \( \Delta x \) is along this interval. This is \( \Delta x \) from A to B. This is really just going to be the initial velocity, which is really just the steady velocity throughout the whole problem because there's no acceleration in the x-axis. So in other words, it's just \( b_x, v_{ax} \times \Delta t \) from A to B. Okay? So we actually know what this \( \Delta x \) is. We also know what the initial velocity is in the x-axis, so we can really easily solve for this \( t_{ab} \). Okay? So, really, what this is going to be is just going to be \( \Delta x \) from A to B divided by \( v_x \). That's going to equal \( t_{ab} \). And that's really just going to be the 10 meters that it travels in the x-axis divided by its velocity in the x-axis, which we just figured out over here was just 4. 4 meters per second positive. Right? It's along the x-axis. So this is really just going to be 2.5 seconds. And that means that now that we figured out what this \( t_{ab} \) is, you can bring it back into this equation over here because we know this is 2.5. Okay? So what is now all we have to do is really just figure out which equation do we use to solve, and it's actually just going to be the third one. It's going to be the one that ignores the final velocity, which is always going to be that third equation. Okay? So what I'm going to do here is now that I'm done, I'm going to pick equation number 3. So \( \Delta y \) from A to B, which is exactly what I'm looking for over here, is just going to be the initial velocity, \( v_0yt_{ab} \), plus one half. This is going to be \( a_y \), and this is going to be \( t_{ab}^2 \). So, in other words, this is really just going to be my initial velocity, which is negative 3. So this is negative 3 times the \( t_{ab} \), which is 2.5 plus 1 half. This is going to be negative 9.8, and then this is going to be 2.5 squared again. When you plug all of this in, just be really careful. Remember, you have negative signs here and here. What you're actually going to see here is that this is equal to negative 38.1 meters. So in other words, the displacement of the rock from the rooftop down to the ground is negative 38.1 meters.

Now, what we said is you could probably just leave your answer like this. But if your professors sort of like a stickler about signs and just wants you to pay attention to the signs very carefully, what you could say is that the height of the building, which is really just the absolute value of \( \Delta y \) from A to B, is actually just 38.1 meters. So either way you represent this answer, this is probably the more technically correct one, but that's just sort of a small point there. Alright? So that's really it for this one. Let me know if you have any questions, and let's get some practice.

Water pours from a spout at the end of a gutter with a speed of 3.5 m/s, where the spout is angled 45° downwards. The magnitude of the water's velocity when it hits the ground is 14 m/s. How high is the spout from the ground?

• A. 18.8 m above the ground
• B. 10.6 m above the ground
• C. 10.0 m above the ground
• D. 9.4 m above the ground

Cannon Firing Downward

Guys, let's check out this problem here. So we've got a cannon that's firing a projectile with speed and angle; we're told that it's 49 degrees below the horizontal. What that means is that it's going to be a downward launch problem. So let's just go ahead and stick to the steps. First, we need to draw a diagram, draw our path in the x and y axes. So we've got a cannon that's fired, and then it's eventually going to move downwards and hit the ground. So our cannonball will look something like this. We're told that this initial velocity is going to be downwards. We know that the angle relative to the horizontal is negative 49 degrees, and then it's just going to take a downward parabolic path until it hits the ground. In the x axis, if you were to move along the x only, it would look like this; in the y axis, it would look like this. So where are points of interest? Well, it's always just the initial, and then what happens is it's just going to hit the ground later, that's the only other point of interest that we're told, so that's just point b. So that means our path in the x and y looks just like this. So that's the first step. What are we actually looking for? That's the second step. What's the target variable? We're looking for the horizontal distance that the cannonball travels before it hits the ground. So what we're looking for here is the horizontal distance covered between points a and b. So this actually takes care of steps 2 and 3. We have the variable and also the interval. We're looking for the interval from a to b because we're looking for the distance between the point where it's fired, and then the point where it hits the ground. In the x-axis, we only have one equation, Δx from a to b = v_x times t. So we're looking for Δx. I'm going to need my initial velocity in the x axis, which I can get because I have the magnitude, I have the v₀ is 73, and I have the angle. So my v_0x, which is v_ax, which is just v_x throughout the whole thing, is going to be 73 times the cosine of negative 49 degrees. We go ahead and plug this into your calculator with the negative sign, then you're just going to get 47.9, and then your initial velocity in the y axis, v_ay, is going to be 73 times the sine of negative 49 with the negative sign, and you're going to get negative 55.1. The reason it's negative is that the initial velocity in the y axis points downwards. So that makes sense. So, I mean, we have these v_x components. The one thing we don't have, though, is we don't have the time. We don't know how long it takes to get from a to b. So we're stuck here, and so we're going to need to go to the y axis to solve. We need my 3 out of 5 variables. The first one we always start with is a_y, which we know is negative 9.8. Our v_0y which is v_ay, we just figured out, is negative 55.1. The final velocity is going to be the velocity at point b, so that's going to be basically what's the y component of the velocity here, which we don't know. Then we have Δy from a to b, and then we have t from a to b. Remember, we came here because we were looking for time so that we can plug it back into this equation here. So we're going to need one out of these variables, we just need only one more. So between v_by and Δyab, which one do we know? Well, let's see. We're told that the fort is 300 meters above the ground, which means that the vertical displacement from a to b as we're going along this path here, this horizontal displacement we know is 300, but because we're going from top to bottom then that means this is going to be negative 300. So we've got our 3 out of 5 variables, 1, 2, 3. So we're going to pick an equation that ignores this final velocity here. So let's go ahead and do that. And, that equation that ignores the final velocity is going to be equation number 3. So that is going to be the equation that we use. So we're going to use equation number 3, which is Δy from a to b is equal to v_ayt_a+^1/2 a_y t²_ab. Remember, I came over here looking for the times, that's what we're really solving for. So we have Δy from a to b, we have v_ay, and we have a_y, so we just plug in everything. Right? So this is going to be negative 300 equals negative 55.1 times t from a to b, plus one half negative 9.8 times t_ab². So these are all just numbers, and the one variable that remains here is t_ab. How do we solve for this? Well, if you look at this equation here, we can rearrange this because we've got some terms of t_ab and then we've also got some terms of t_ab². I'm going to rearrange this equation. First of all, I'm going to do one half times negative 9.8, and I'm going to put this out in front. So I'm going to do negative 4.9. That's just what happens when you cut negative 9.8 in half. So this is going to be negative 4.9t_ab², and then I'm going to do minus 55.1t_ab, that's this term over here. And then I'm going to move this term to the other side, this negative 300 to the other side, and it becomes positive 300, and that's going to equal 0. The reason I'm going to do this is that if you take a look now, this equation is kind of in the form a times t², plus b times t, plus c, which is a constant equals 0. The reason this is useful for us is that we have these coefficients here, a, b, and c, which are all just these numbers that go before the t's, and the way we solve this kind of equation here is by using the quadratic formula. So we use the quadratic formula sometimes to solve your projectile motion questions. And basically, the quadratic formula says that if you want to solve for t, your t is going to have 2 solutions. It's going to be negative b remember you probably learned a song for this, plus or minus the square root of b² minus 4ac all over 2a. So what happens is if you just plug in all of these numbers into this formula, then you're just going to get 2 solutions for t. I'm going to go ahead and work this out for you guys. So negative b. Well, b in this case is going to be negative 55.1. So the negative of a negative is going to be a positive 55.1, plus or minus the square roots of negative 55.1² minus 4 times a, which is negative 4.9. Keep track of the negative signs there, and this is going to be 300. Okay. So now we're going to do all this divided by 2 times negative 4.9. Okay. So if you go ahead and work this out in your calculator, the plus and minus case, then what you're going to get is you're going to get 2 solutions for time. You're going to get negative 15 and then you're going to get 4.0 seconds. Remember, because one these solutions won't make sense, and this one doesn't make sense because it doesn't make sense to have a negative time. So it doesn't make sense. Okay. So that means our time here is actually 4.0 seconds, which is great because remember we needed time to plug it back into our x equation to figure out what the Δx is, the horizontal displacement. So we actually do that down here. I'm going to say that Δx_ab is equal to v_ax, v_x times t_ab. And so Δx is just going to equal the horizontal velocity, which is 47.9 times 4.0. And if you go ahead and work this out, what you're going to get is you're going to get 191.6, which can be rounded to 192 meters, and so that is your answer for the horizontal displacement. So sometimes you may have to use the quadratic formula inside of these cases here. So, you know, hopefully, be on the lookout for that. Anyway, so that means our answer choice is going to be answer choice b. Let me know if you guys have any questions. Let's keep going.

Here’s what students ask on this topic:

What is the initial vertical velocity in a downward launch?

In a downward launch, the initial vertical velocity (v_0y) is negative because the chosen positive direction is upwards. For example, if an object is thrown downward at an angle, you can calculate v_0y using the equation:

$v_{0}y=v_{0}sin(\theta )$

where θ is the angle of launch measured from the horizontal. If θ is negative (indicating a downward direction), v_0y will also be negative.

How do you calculate the height of a building in a downward launch problem?

To calculate the height of a building in a downward launch problem, you use the vertical displacement equation:

$\mathrm{\Delta y}=v_{0}yt+\frac{1}{2}{a}_y{t}^2$

Here, $v_{0}y$ is the initial vertical velocity, $a_y$ is the acceleration due to gravity (−9.8 m/s²), and $t$ is the time of flight. The absolute value of $\mathrm{\Delta y}$ gives the height of the building.

Why is the initial vertical velocity always negative in a downward launch?

The initial vertical velocity (v_0y) is always negative in a downward launch because the positive direction is defined as upwards. When an object is launched downward, its initial velocity vector points in the opposite direction of the positive y-axis. Therefore, v_0y is negative to reflect this downward direction.

What equations are used to solve downward launch problems?

To solve downward launch problems, you use the same kinematic equations as for any projectile motion. Key equations include:

1. $\mathrm{\Delta y}=v_{0}yt+\frac{1}{2}{a}_y{t}^2$

2. $\mathrm{\Delta x}=v_{0}xt$

3. $v_y=v_{0}y+a_{y}t$

Here, $v_{0}y$ and $v_{0}x$ are the initial velocities in the y and x directions, respectively, and $a_y$ is the acceleration due to gravity.

How do you determine the time of flight in a downward launch problem?

To determine the time of flight in a downward launch problem, you can use the horizontal motion equation:

$\mathrm{\Delta x}=v_{0}xt$

Here, $\mathrm{\Delta x}$ is the horizontal displacement, and $v_{0}x$ is the initial horizontal velocity. Rearrange the equation to solve for time:

$t=\frac{\mathrm{\Delta x}}{v_0}$

Once you have the time, you can use it in the vertical motion equations to find other variables.